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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
∠B = ∠A + d as A, B, C, are in AP
∠C = ∠A + 2d
∠A + ∠B + ∠C = 180°
∠A + ∠A + d + ∠A + 2d = 180°
3∠A + 3d = 180°
∠A + d = 60°
∠B = ∠A + d = 60°
sin (2A + B) = 1/2
sinθ = 1/2
θ = 30° or θ = 150°
Since B = 60° & A can’t be negative so
θ ≠ 30° hence θ = 150°
2A + B = 150°
2A + 60 = 150
2A = 90°
∠A = 45°
∠B = ∠A + d
60 = 45 + d
d = 15°
∠C = 45 + 2 × 15°
∠C = 75°
sin (60 + 2 × 75) = sin( 60 + 150)
sin210° = sin(180 + 30)
= - sin30° = - 1/2